Problem of the Week #43 - March 25th, 2013

[Chris L T521]

Here's this week's problem.

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Problem: Use contour integration to show that
\[\int_{-\infty}^{\infty}\frac{e^{-2\pi i x\xi}}{(1+x^2)^2}\,dx = \frac{\pi}{2}(1+2\pi|\xi|)e^{-2\pi|\xi|} \]
for all $\xi$ real.

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Hint:

Spoiler

WLOG, suppose that $\xi\geq 0$ (this way, you don't have to worry about the absolute values for the time being). Then consider using the lower half circle as the contour for this integral.

Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Chris L T521]

No one answered this week's question. Here's my solution below.

Spoiler

WLOG, suppose that $\xi\geq 0$ (so we don't have to worry about absolute values for the time being). Let the closed contour $\Gamma$ be the lower half circle. Let $f(z)=\frac{e^{-2\pi i z\xi}}{(1+z^2)^2}$. Clearly, $f(z)$ has two poles of order $2$ at $z=i$ and $z=-i$. Based on how we defined $\Gamma$, $z=i$ is not contained within the closed contour, so we need to find the residue at $z=-i$. We see that\[\begin{aligned}\text{res}_{-i}f(z) &= \lim_{z\to-i}\frac{\,d}{\,dz}\left[\frac{e^{-2\pi i z\xi}}{(z-i)^2}\right]\\ &=\lim_{z\to-i}\frac{-2\pi i \xi(z-i)^2e^{-2\pi i z\xi}-2(z-i)e^{-2\pi iz\xi}}{(z-i)^4}\\ &=-i\frac{(1+2\pi\xi)e^{-2\pi\xi}}{4}\end{aligned}\]
Therefore, by the residue theorem, we see that
\[\int_{\Gamma}\frac{e^{-2\pi iz\xi}}{(1+z^2)^2} = \int_{-R}^{R}\frac{e^{-2\pi i x\xi}}{(1+x^2)^2}+\int_{C_R}\frac{e^{-2\pi iz\xi}}{(1+z^2)^2} = \frac{\pi}{2}(1+2\pi\xi)e^{-2\pi\xi}\]
Observe that $e^{-2\pi iz\xi}\leq 1$ for $\text{Im}z\leq 0$ and $\xi\geq 0$. Thus, as $R\rightarrow\infty$,
\[\left|\int_{C_R}\frac{e^{-2\pi iz\xi}}{(1+z^2)^2}\,dz\right|\leq \frac{\pi R}{(R^2+1)^2}\]
which goes to zero. Therefore, as $R\rightarrow\infty$, we have
\[\int_{-\infty}^{\infty}\frac{e^{-2\pi ix\xi}}{(1+x^2)^2}\,dx=\frac{\pi}{2}(1+2\pi|\xi|)e^{-2\pi|xi|}\]
and the proof is complete.

(PS: I just finished posting a solution to POTW 42, so feel free to check that out too!)